CHNG2806/9206 Separation Processes
Membrane separation and transport phenomena in membrane processes
Week 9, Tutorial, solutions
1. The following solutions have concentration of 100 mg/L. Calculate their osmotic pressure at
25 oC, respectively.
(1) Sucrose (C12H22O11) solution;
(2) KCl solution;
(3) Na2SO4 solution.
(hints: note the concentration unit; check the molecular weight for each solution; determine the
van’t Hoff coefficient for each solute)
Solution:
Use van’t Hoff equation to calculate the osmotic pressure.
Data
R = 0.083145 L bar mol-1K-1
𝜋 = 𝐶𝛽𝑅𝑔 𝑇
Solute
Sucrose
KCl
Na2SO4
Ci(mg/L)
100
100
100
MW (g/mole)
342
74.5
142
Ci (mole/L)
2.924 × 10-4
1.342 × 10-3
7.042 × 10-4

1
2
3
 (bar)
7.244 × 10-3
6.652 × 10-2
5.235 × 10-2
2. Use the solution-diffusion model in osmotically driven membrane processes (i.e., forward
osmosis (FO) and pressure-retarded osmosis (PRO))
(1) Derive the specific reverse solute flux (Js/Jw) for osmotically driven membrane processes
using solution-diffusion model.
(2) An FO membrane has a A value of 2 LMH/bar and B value of 0.2 LMH for NaCl. Calculate
the Js/Jw at 25 oC.
(3) A PRO membrane has a A value of 2 LMH/bar and B value of 0.2 LMH for NaCl. Calculate
the Js/Jw at an applied hydraulic pressure of 10 bar at 25 oC when Jw is 25 LMH.
Solution:
(1) The derivation is below.
Based on the solution-diffusion theory for the water and solute transport in the dense rejection
layer, the water flux ( J w ) and the solute flux ( J s ) in PRO processes can be described by:
J w = A( − P)
(A1)
and
J s = BC
(A2)
where A and B are the water permeability coefficient and solute permeability coefficient of
the rejection layer, respectively;  and C are the effective osmotic pressure difference and
solute concentration difference across the rejection layer, respectively; and P is the effective
hydraulic pressure applied in draw solution.
Rearranging (A1)
J w = A( − P)
 = J w / A + P
Substituting
∆𝜋 = 𝛽𝑅𝑔 𝑇. ∆𝐶
𝛽𝑅𝑔 𝑇. ∆𝐶 =
𝐽𝑤
+ ∆𝑃
𝐴
Substituting
C = J s / B
𝛽𝑅𝑔 𝑇.
𝐽𝑠
𝐽𝑤
=
+ ∆𝑃
𝐵
𝐴
Rearranging
𝐵
𝐽𝑤
𝐵 𝐽𝑤
𝐴
( + ∆𝑃) =
( 1 + ∆𝑃)
𝐽𝑠 =
𝛽𝑅𝑔 𝑇 𝐴
𝛽𝑅𝑔 𝑇 𝐴
𝐽𝑤
Js
B
AP
=
(1 +
)
J w A RgT
Jw
(A6)
(2) Use the equation
In FO, ∆𝑃 = 0
Js/Jw = 0.2 LMH / (2 LMH/bar × 2 ×0.083145 L bar mol-1K-1 × 298 K) = 0.002018 mol/L
(3) Use the equation
Js
B
AP
=
(1 +
)
J w A RgT
Jw
Js/Jw = 0.2 LMH / (2 LMH/bar × 2 ×0.083145 L bar mol-1K-1 × 298 K) × (1 + 2 LMH/bar
× 10 bar / 25 LMH) = 0.003632 mol/L
3. In an RO membrane testing module, feed channel is a spacer-filled channel and feed solution
is NaCl. Calculate the water flux for the following conditions:
Feed concentration 0.5 M NaCl, apparent rejection 99%, membrane water permeability
2.5 LMH/bar, K, mass transfer coefficient, 8.9×10-5 m/s, and applied hydraulic pressure
10 bar.
(hints: need to first find/derive the water flux equation incorporating concentration
polarization; iterative calculation method may be used for water flux calculation)
Solution:
From the rejection equation,
Osmotic Pressure Drop
∆𝜋 = 𝛽𝑅𝑇∆𝐶 = 2𝑥0.08314 ∗ (25 + 273) ∗ (0.5 − 0.005) = 24.52 𝑏𝑎𝑟𝑠
Water Flux (+ CP)
Data consistency (convert all to SI)
𝐴 = 2.5
𝐿
𝑚3
= 6.94𝑥10−7 2
𝑚2. ℎ. 𝑏𝑎𝑟
𝑚 . 𝑠. 𝑏𝑎𝑟
𝐽𝑤 = 𝐴 (∆𝑃 −
𝐽
( 𝑤)
𝑒 𝑘𝑠 ∆𝜋)
𝐽𝑤 = 2.5𝑥10−7 (10 − 𝑒
Jw = -1.0x10-3 (m3/m2.s) or -360 LMH
Flux is negative because P < 
(
𝐽𝑤
)
8.9𝑥10−5 24.52)
4. A process engineer tested a RO membrane performance. When the feed was pure water and
the applied hydraulic pressure was 5 bar, she measured a membrane water flux of 25 LMH.
Then she added some NaCl into the feed to reach a concentration of NaCl 10 mM in the feed
solution. After stabilized, she measured the permeated concentration was 0.1 mM and the water
flux dropped to 22 LMH at the same applied hydraulic pressure. Calculate:
(1) membrane water permeability;
(2) membrane salt permeability;
(3) concentration polarization factor under the current testing conditions.
(hints: first find/derive the equation for calculating A and B values)
Solution:
(1) Q4
4.3
𝐽𝑤
𝐽𝑤 = 𝐴(∆𝑃 − ∆𝜋𝑚 ) = 𝐴[∆𝑃 − (𝜋𝑚 − 𝜋𝑝 )] = 𝐴 [∆𝑃 − 𝑒𝑥𝑝 ( ) (𝜋𝑏 − 𝜋𝑝 )]
𝐾
Based on vant’s Hoff equation
𝜋𝑏 − 𝜋𝑝 = (𝐶𝑏 − 𝐶𝑝 )𝛽𝑅𝑔 𝑇 = (10 − 0.1) × 10−3 × 0.083145 × 298 = 0.4906 𝑏𝑎𝑟
Jw = 22 LMH, so:
22 = 5[5 − 𝑓𝑐𝑝 × 0.4906]
𝑓𝑐𝑝 = 1.22
Q4 (2) At steady state rate of solute flux equal the water flux x concentration of permeate
𝐽𝑠 = 𝐵∆𝐶 = 𝐽𝑤 𝐶𝑝
∆𝜋
∆𝐶 =
𝛽𝑅𝑔 𝑇
𝐵 = 𝐽𝑤 𝐶𝑝
𝛽𝑅𝑔 𝑇
2𝑥0.08314𝑥(25 + 273)
= 22𝑥0.10−3
= 0.219 𝐿𝑀𝐻
∆𝜋
0.49
5. A student was doing RO membrane performance test. He measured a pure water flux of 25
LMH at an applied hydraulic pressure of 10 bar. After adding some NaCl to the feed solution
to reach a concentration of 20 mM, he measured the permeate NaCl concentration of 1 mM
and the water flux declined to 21 LMH at stabilized condition. After adding a certain amount
of colloidal silica for fouling test, the flux further declined to 8 LMH and the permeate NaCl
concentration increased to 2 mM. He also detected that the foulant resistance on the membrane
was half of the clean membrane resistance. Calculate:
(1) Clean membrane resistance;
(2) CP factor when the feed only contained 20 mM NaCl;
(3) CP factor during the fouling test;
(4) Briefly explain why the CP factor was increased in the fouling test.
(assume that the feedwater viscosity is 8.90×10-4 Pa.s and is unchanged under all the
experimental conditions)
Note: water flux unit LMH means Lm-2h-1
Change all the units to SI
Conditions
No salt
CP
Fouling
Jw
Pressure
(P)
6.94x10-6 m3/ m2.s
5.83 x10-6
2.22 x10-6
106 Pa
10 bar
No salt
CP
Fouling
Conc

Rg
T
i)
25 L/m2.h
21
8
20 mM
1 mM
2 mM
20
1
2
-4
8.90x10
8.314
298
mole/m3
mole/m3
mole/m3
Pa.s
m3.Pa/moleK
K
No salt
Water Flux
𝐽𝑤 = 𝐴(∆𝑃 − ∆𝜋) =
Since  = 0
(∆𝑃 − ∆𝜋)
𝜇𝑅𝑚
𝐽𝑤 = 𝐴(∆𝑃) =
𝑅𝑚 =
(∆𝑃)
6.94𝑥10−6
1
=
= 1.621014 ( )
𝜇𝐽𝑤
8.90x10 − 4x6.94x10 − 6
𝑚
𝐴=
ii)
(∆𝑃)
𝜇𝑅𝑚
𝐽𝑤
6.94x10 − 6
𝑚3
=
= 6.941012 3
(∆𝑃)
106
𝑚 . 𝑠. 𝑃𝑎
CP
∆𝜋 = 𝛽𝑅𝑔 𝑇∆𝐶 = 2.0𝑥8.314x298x(20 − 1) = 94148 Pa
Water Flux
𝐽𝑤 = 𝐴(∆𝑃 − 𝑓𝐶𝑃 ∆𝜋) =
𝑓𝐶𝑃
iii)
(∆𝑃 − 𝑓𝐶𝑃 ∆𝜋)
𝜇𝑅𝑚
5.83 x10 − 6
𝐽
)
(∆𝑃 − 𝐴𝑤 ) (106 −
6.941012
=
=
= 1.70
∆𝜋
94148
Fouling
∆𝜋 = 𝛽𝑅𝑔 𝑇∆𝐶 = 2.0𝑥8.314x298x(20 − 2) = 89193 Pa
Since Rf = 0.5 Rm
𝐽𝑤 =
𝑓𝑓,𝐶𝑃 =
(∆𝑃 − 𝑓𝑓,𝐶𝑃 ∆𝜋) (∆𝑃 − 𝑓𝑓,𝐶𝑃 ∆𝜋)
=
𝜇[𝑅𝑚 + 𝑅𝑓 ]
𝜇1.5[𝑅𝑚 ]
(∆𝑃 − 𝜇1.5[𝑅𝑚 ]𝐽𝑤 ) (106 − 8.90x10 − 4x1.5x1.621014 𝑥2.22 x10 − 6)
=
= 5.83
∆𝜋
89193